Continous function, a interesting view

We are familiar with the continuous function on $\mathbb{R}$$\mathbb R$. At the beginning, I think it is trivial, But obviously, it is not.

For example, consider the algebraic structure of $C$$C$,

We can define $(f+g)(x)=f(x)+g(x),(fg)(x)=f(x)g(x),(\lambda f)(x)=\lambda f(x)$$(f+g)(x)=f(x)+g(x),(fg)(x)=f(x)g(x),(\lambda f)(x)=\lambda f(x)$

This shows that $C$$C$ is an Algebra, I mean both linear space and ring.

And actually, $a(f):=f(a)$$a(f):=f(a)$ is an Algebra homomorphism!

$a(f):C\to \mathbb{R}$$a(f):C\to \mathbb R$ $a(f+g):=(f+g)(a)=f(a)+g(a)=a(f)+a(g)$$a(f+g):=(f+g)(a)=f(a)+g(a)=a(f)+a(g)$

$a(fg)=fg(a)=f(a)g(a)=a(f)a(g)$$a(fg)=fg(a)=f(a)g(a)=a(f)a(g)$

$a(\lambda f)=\lambda f(a)=\lambda a(f)$$a(\lambda f)=\lambda f(a)=\lambda a(f)$

By the way, the limit is also an Algebra homomorphism for convergence sequence.

And consider $a\in \mathbb{R}$$a\in \mathbb R$, all the $f\in C,a(f):=f(a)=0$$f\in C,a(f):=f(a)=0$ give an idea $\mathcal{I}$$\mathcal I$ of $C$$C$

${\displaystyle \frac{C}{\mathcal{I}}\cong \mathbb{R}}$$\displaystyle\frac{C}{\mathcal I}\cong\mathbb R$ by the first isomorphism theorem.

The topological structure is also interesting,

It is a complete metric space by considering $(C_{[a,b]},d_{\mathrm{\infty }})$$(C_{[a,b]},d_{\infty})$

$d_{\mathrm{\infty }}(f,g)=sup\parallel f(x)-g(x)\parallel ,x\in [a,b]$$d_{\infty}(f,g)=\sup\| f(x)-g(x)\|,x\in [a,b]$

So consider a Cauchy sequence, $\mathrm{\forall }ϵ>0,\mathrm{\exists }\mathbb{N},\mathrm{\forall }n,m>N,sup\parallel f_n(x)-f_m(x)\parallel \le ϵ$$\forall \epsilon>0,\exists\N,\forall n,m> N,\sup\|f_n(x)-f_m(x)\|\le\epsilon$

We need to prove that it is convergence.

Consider $sup\parallel f(x)-f_n(x)\parallel =sup\parallel f(x)-f(x-\delta )+f(x-\delta )-f_n(x-\delta )+f_n(x-\delta )-f_n(x)\parallel$$\sup\| f(x)-f_n(x)\|=\sup\|f(x)-f(x-\delta)+f(x-\delta)-f_n(x-\delta)+f_n(x-\delta)- f_n(x)\|$

$\begin{array}{c}sup\parallel f(x)-f(x-\delta )+f(x-\delta )-f_n(x-\delta )+f_n(x-\delta )-f_n(x)\parallel \\ \le sup\parallel f(x)-f(x-\delta )\parallel +sup\parallel f(x-\delta )-f_n(x-\delta )\parallel +sup\parallel f_n(x-\delta )-f_n(x)\parallel \le \frac{ϵ}{3}+\frac{ϵ}{3}+\frac{ϵ}{3}=ϵ\end{array}$$\sup\|f(x)-f(x-\delta)+f(x-\delta)-f_n(x-\delta)+f_n(x-\delta)- f_n(x)\|\\\le\sup\|f(x)-f(x-\delta)\|+\sup\|f(x-\delta)-f_n(x-\delta)\|+\sup\| f_n(x-\delta)-f_n(x)\|\le \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$

And we can consider a new view for $\begin{array}{c}{\int }_a^{b}f(x)dx\end{array}$$\int_a^b f(x) dx\\$

It is a linear continuous functional !,(I think it belongs to the dual space of $C$$C$)

To see it is continuous, we can consider that $d_{\mathrm{\infty }}(f,g)\le \delta$$d_\infty(f,g)\le\delta$

$\begin{array}{c}{\int }_a^{b}f(x)-g(x)dx\le {\int }_a^b|f(x)-g(x)|dx\le {\int }_a^b\delta dx=\delta (b-a)\end{array}$$\int_a^b f(x)-g(x) dx\le\int_a^b|f(x)-g(x)|dx\le\int_a^b\delta dx=\delta(b-a)\\$

So $\begin{array}{c}\mathrm{\forall }ϵ>0,\mathrm{\exists }\delta =\frac{ϵ}{b-a},d_{\mathrm{\infty }}(f,g)\le \delta \Rightarrow {\int }_a^{b}f(x)-g(x)dx\le ϵ\end{array}$$\forall\epsilon>0,\exists \delta=\frac{\epsilon}{b-a},d_{\infty}(f,g)\le\delta\Rightarrow\int_a^b f(x)-g(x) dx\le\epsilon\\$

So because of $\begin{array}{c}{\int }_a^{b}f(x)dx\end{array}$$\int_a^bf(x)dx\\$ is continuous, $\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim}{\int }_a^b{f}_n(x)dx={\int }_a^{b}f(x)dx\end{array}$$\lim_{n\to\infty} \int_a^bf_n(x)dx=\int_a^bf(x)dx\\$